What is the antiderivative of #xe^x#?

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Answer 1 · 2016-07-08T07:39:11

Let #u=x# and #dv= e^(x) dx#, so #du=dx# and #v=e^(x)#

Knowing that

#int u dv = uv - int v du#

You can see that

#int x e^(x) dx#
#= xe^(x)-int e^x dx #
#= xe^(x)-e^(x)+C#

#=e^(x)(x-1)+C#

In this case, you'd have to perform integration by parts, which is given by the following formula:

#int u dv = uv - int v du#

Note that if you would have chosen #u=e^(x)# and #dv = x dx#, your integral would be even more complicated.